How Does Mass Transfer Between Rotating Drums Affect Their Angular Velocities?

[Order]

Here is a problem I spent my sunday trying to solve. I made progress, but still something is wrong.

Homework Statement

A drum of mass M_A and radius a rotates freely with initial angular velocity \omega_{A}(0). A second drum with mass M_B and radius b>a is mounted on the same axis and is at rest, although it is free to rotate. A thin layer of sand with mass M_S is distributed on the inner surface of the smaller drum. At t=0 small perforations in the inner drum are opened. The sand starts to fly out at a constant rate \lambda and sticks to the outer drum. Find the subsequent angular velocities of the two drums \omega_{A} and \omega_{B}. Ignore the transit time of the sand.
Ans. clue. If \lambda t = M_{B} and b=2a then \omega_{B}=\omega_{A}(0)/8

Homework Equations

\frac{dL}{dt}=0

L=mr^{2}\omega

The Attempt at a Solution

I begin by looking at the angular momentum of the inner drum: L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}\omega_{A}(t) and at a later time L_{A}(t+\Delta t)=(M_{A}+M_{S}-\lambda (t+\Delta t))a^{2}\omega_{A}(t+\Delta t)+\lambda \Delta t a^{2}\omega_{A}(t). Taking the limit and dividing by dt gives \frac{dL_{A}}{dt}=(M_{A}+M{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}-\omega_{A}\lambda a^{2}=0. This is a differential equation which can be solved by switching a little to give
\omega_{A}=\omega_{A}(0)\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}
I think this might be right.

However, the second part I am not so sure of. First of all some energy is lost when the sand falls in partly perpendicular to the surface of the outer drum. One can show that a fraction a/b of the momentum survives and the rest becomes heat. Therefore the angular momentum of the outer drum at time t will be L_{B}(t)=(\lambda t + M_{B})b^{2}\omega_{B}(t)+b\frac{a}{b}a\omega_{A} \lambda \Delta t, where the second term is the incoming sand losing some mechanical energy. At a later time L_{B}(t)=(\lambda (t+\Delta t) + M_{B})b^{2}\omega_{B}(t+\Delta t). Taking the limit and dividing by dt gives \frac{dL_{B}}{dt}=(\lambda t+M_{B})b^{2}\frac{d\omega_{B}}{dt}+\lambda(\omega_{B}b^{2}-\omega_{A}(0)a^{2}\frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t})=0 This is a linear differential equation which can be solved to give \omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}}\ln \lgroup \frac{M_{A}+M_{S}}{M_{A}+M_{S}-\lambda t}\rgroup \frac{M_{A}+M_{S}}{M_{B}+\lambda t}
This is not in accordance with the clue, nor with my answer sheet that says \omega_{B}=\omega_{A}(0)\frac{a^{2}(M_{A}+M_{S}-\lambda t)}{b^{2}(M_{B}+\lambda t)}
Please tell me if something is unclear.

 

[RTW69]

Isn't your first differential equation dWa/Wa=lambda*dt/(Ma+Ms-lambda*t)?

 

[Order]

Isn't your first differential equation dWa/Wa=lambda*dt/(Ma+Ms-lambda*t)?

Yes it is \frac{d\omega_{A}}{\omega_{A}}=\lambda \frac{dt}{M_{A}+M_{S}-\lambda t} This leads to \ln \omega_{A}=-\ln (M_{A}+M_{S}-\lambda t)+C which leads to my equation stated in original post.

However I think that the angular momentum of the inner drum might be constant. This is because now looking closer at my equations L_A(t) and L_A(t+delta*t) I get a sign difference. L_{A}(t+\Delta t)-L_{A}(t)=(M_{A}+M_{S}-\lambda t)a^{2}(\omega_{A}(t+\Delta t)-\omega_{A}(t))+\lambda a^{2}\Delta t(\omega_{A}(t)-\omega_{A}(t+\Delta t)) Now the second term in this equation becomes zero when taking the limit, so (M_{A}+M_{S}-\lambda t)a^{2}\frac{d\omega_{A}}{dt}=0 with the only solution that the angular momentum is constant.

Now putting this into the differential eguation for the second drum \frac{d\omega_{B}}{dt}+\frac{\lambda}{\lambda t+M_{B}}\omega_{B}=\omega_{A}(0)\frac{a^{2}}{b^{2}}\frac{\lambda}{\lambda t+M_{B}}, with the solution \omega_{B}(t)=\omega_{A}(0)\frac{a^{2}}{b^{2}} \frac{\lambda t}{\lambda t+M_{B}} This is correct according to the answer clue but not according to my answer sheet. Do you think this might be right? Will a drum losing mass have a constant angular momentum?